You are an investigator at the scene of a car crash. A Volvo of has hit an Alpha

Question

You are an investigator at the scene of a car crash. A Volvo of has hit an Alpha Romeo from the side and the two are inter-tangled (stuck together). You can see that the Volvo was traveling due east on a road with a 50 mph speed limit and made a 30.0 m long skid mark before reaching the intersection where it hit the Alpha Romeo which was driving due North on a road with a 25 mph speed limit. There are additional skid marks from both cars that are 12.25 m long at 15 degrees North of East. Further investigation tells you that the mass of the Volvo (including the driver) is 1650 kg and the mass of the Alpha Romeo (including the driver) is 975 kg. Your task is to determine if either car was speeding.

1. Determine the speed of the Volvo before it slammed on the brakes.

2. Find the speed of the Alpha Romeo when it was hit by the Volvo.

Explanation / Answer

let speed of volvo when slamming brake = v, Mass of volvo = m1

speed of volvo just before coliision = v1

After breaking , skidding distance = 30 m

v1^2- v^2 = - 2 X a X s , a= deceleration and s= skidding distance

Let us assume that the dceleration = -0.3 g = - 2.94 m/s^2 ( Taken from Volvo web-site - for car's emergency system)

v^2 = v1^2 + (2 X a X s )= v1^2 + 176.4

v^2 = v1^2 + 176.4

After collision both cars coalesc and travel together ( m1+m2) with intial speed of V ( say)

Then the combined cars travel 12.25 mrs and come to halt

(0)^2 -(V)^2 = - 2 X a' X s' , where a' = deceeration of both cars combined , s' = distance travelled by them

Let us assume the deceleration of combined system is higher at (- g )

V^2 = 2 X 9.8 X 12.25

V = 15.5 m/s

X- direction ( East) m1 X v1 = (m1+ m2) X V X cos 15 ------ conservation of momentum

v1 = (1650+ 975) Kg X 15.5 m/s X cos 15 /1650 Kg = 23.8 m/s

So v^2= V1^2 + 176.4 = (23.8)^2 + 176.4

v = 27.3 m/s = 61.1 mph ( He is driving beyond speed limit while hitting the brakes )

m2 X v2 = (m1+m2) V sin 15 ------ conservation of momentum for Y-direction

v2 = (m1+m2) X V X sin 15/ m2

v2 = 41.7 m/s ( 93.3 m/hr ) - exceeding speed limit

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